Section 6 – 4D:
Finding a Value of x with a Given tail arae
Label the shaded area for both graphs. Find the value for z and label the z axis. Find the value
for x for the given area under the normal curve with the specified mean and standard deviation.
Round the value for x to 2 decimal places.
Example 1
Given: µ x = 3.2 and σ x = .45
Find the x value that has a left tail area of .1131
If the left tail area for the x value
is .1131
P ( x < ??? ) = . 1131
then the left tail area for the z curve
is also .1131
left tail area left tail area = .1131
= .1131
x < ???
µ x = 3.2
σ x = .45
X
z < ???
Z
0
1
Find a Z score that has a left tail area as close to .1131 as possible.
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–1.2
0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020
0.08
0.09
0.1003
0.0985
If Z = –1.21 then convert z to x by the following formula x = µ + z(σ )
setup for finding x:
x = µ + z(σ )
x = 3.2 –1.21(.45) = 2.66
x = 2.66
P ( x < 2.66 ) = .1131
English conclusion:
About 11% of the x values are less than 2.66
Stat 300 6 – 4D Lecture
Page 1 of 5
© 2012 Eitel
Example 2
Given:
µ x = 21.9 and σ x = 3.4
Find the x value that has a right tail area of .1134
If the right tail area for the
x distribution is .1314
then the left tail area is .8686
P ( x > ??? ) = . 1314
If the left tail area for the
x distribution is .8686 then the
left tail area for the z curve is also .8686
Ieft tail area right tail area Ieft tail area = .8686
= .1314
= .8686
µ x = 21.9
σ x = 3.4
x = ???
X
0
1
Z
z = ???
Find a Z score that has a left tail area as close to .8686 as possible.
Positive Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
1.1
0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790
0.08
0.09
0.8810
0.8830
If Z = 1.12 then convert z to x by the following formula x = µ + z(σ )
setup for finding x:
x = µ + z(σ )
x = 21.9 + 1.12(3.4) = 25.71
x = 25.71
P ( x > 25.71 ) = . 1314
English conclusion:
About 13% of the x values are more than than 25.71
Stat 300 6 – 4D Lecture
Page 2 of 5
© 2012 Eitel
Example 3
Given:
µ x = 92.6 and σ x = 7.1
Find the x value that has a right tail area of .9382
If the right tail area for the
x distribution is .9382
then the left tail area is .0618
P ( x > ??? ) = . 9382
If the left tail area for the
x distribution is .0618 then the
left tail area for the z curve is also .0618
Ieft tail area right tail area = .0618
= .9382
Ieft tail area = .0618
x = ???
µ x = 92.6
σ x = 7.1
x
z = ???
Z
0
1
Find a Z score that has a left tail area as close to .8686 as possible.
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–1.5
0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582
0.08
0.09
0.0571
0.0559
If Z = –1.54 then convert z to x by the following formula x = µ + z(σ )
setup for finding x:
x = µ + z(σ )
x = 92.6 −1.54(7.1) = 81.67
x = 81.67
P ( x > 81.67 ) = . 9382
English conclusion:
About 94% of the x values are more than than 81.67
Stat 300 6 – 4D Lecture
Page 3 of 5
© 2012 Eitel
Example 4
Kradle to Kindergarten Daycare Center knows that the ages of the children it cares for are normally
distributed with a mean age of 3.2 years and a standard deviation of .45 years. The daycare center
wants to mail out an advertisement to the parents of the youngest 35% of its members. What is the
age that separates the youngest 35% of the children from the older children?
Given:
µ x = 3.2 and σ x = .45
Find the x value that has a left tail area of .3500
P ( x < .??? ) = . 3500
If the left tail area for the x value
is .3500
then the left tail area for the z curve
is also .3500
left tail area = .3500
left tail area = .3500
x = ???
µ x = 3.2
σ x = .45
X
z = ???
Z
0
1
Find a Z score that has a left tail area as close to .3500 as possible.
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–0.3
0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557
0.08
0.09
0.3520
0.3483
If Z = –.39 then convert z to x by the following formula x = µ + z(σ )
setup for finding x:
x = µ + z(σ )
x = 3.2 – .39(.45) = 3.024
x = 3.02
English conclusion:
If I send an advertisement to the parents of children less than 3.02 years old, the advertisement will
be sent to the youngest 35% of the daycare children.
Note: The age that separates the youngest 35% of the children from the older children is called P35 .
We use a Normal Distribution and the Standard Normal Distribution to find P35 instead of the
sorted list and percentile formula technique used in the past. We do not have the values listed
for the distribution of x values so we cannot sort the list and find the location of P35 .
Stat 300 6 – 4D Lecture
Page 4 of 5
© 2012 Eitel
Example 5
Kradle to Kindergarten Daycare Center knows that the ages of the children it cares for are normally
distributed with a mean age of 3.2 years and a standard deviation of .45 years. The daycare center
wants to send out an add to the parents of the oldest 15% of its members. What is the age that
separates the oldest 15% of the children from the younger children?
Given:
µ x = 3.2 and σ x = .45
Find the x value that has a right tail area of .1500
If the right tail area for the x values
is .1500
then the left tail area
is 1 – .1500 = .8500
P ( x < .??? ) = .1500
If the left tail area for the x values
is .8500
then the left tail area or the z values
is also .8500
left tail area
= .8500
right tail
left tail
area = .15
area = .85
µ x = 3.2 x = ???
σ x = .45
X
Z
z = ???
0
1
If the left tail area for the x values
is .8500
then the left tail area or the z values
is also .8500
Find a Z score that has a left tail area as close to .8500 as possible.
Positive Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
1.0
0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577
0.08
0.09
0.8599
0.8621
If Z = 1.04 then convert z to x by the following formula x = µ + z(σ )
setup for finding x:
x = µ + z(σ )
x = 3.2 + 1.04(.45) = 3.668
x = 3.67
English conclusion:
If I send an add to the parents of children more than 3.67 years old, the add will be sent to the oldest
15% of the daycare children.
Note: The age that separates the oldest 15% from the youngest 15% of the children from the
older children is called P15 .
Stat 300 6 – 4D Lecture
Page 5 of 5
© 2012 Eitel